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**Nuclei**

**Physics**. 438. 13.1 INTRODUCTION. In the previous chapter, we have learnt that
in every atom, the ... various properties of **nuclei** such as their size, mass and
stability, and ... and its constituents, ∆M, is called the **mass defect**, and is given
by ... apart a **nucleus** in this way, the nuclear **binding energy** is still a convenient.

For more information, see **Nuclei**

Given that

Mass of copper coin ^{63}_{29} C, m = 3 g

Atomic mass of copper A = 62.92960 amu

Step 1: Compute the number of atoms given grams and units of atomic mass

Number of atoms = \frac{NA}{A} * \text{ mass }

Where, NA - Avogadro's number ( 6.023 * 10^{23}) atoms/g , A - atomic mass

= \frac{6.023 * 10^{23}}{62.92960} * 3

= 2.868 * 10^{22} atoms

Step 2: Finding the mass defect of all nucleus present in the coin

Number of protons = 29

Number of neutrons = 34

Mass of proton m_H = 1.007825 u

Mass of neutron m_n = 1.008665 u

Mass defect of nitrogen \Delta m = m_H + m_n - m

\Delta m=29*(1.007825)+34*(1.008665)-62.92960

\Delta m=0.592 amu

Mass defect of all nucleus present in the coin \Delta m = 0.592 * 2.868 * 10^{22}

\Delta m = 1.698 * 10^{22} amu

Converting atomic mass units to MeV/c^2

1 amu = 931.5 MeV/c^2

\Delta m=1.698\cdot10^{22}\ \cdot931.5\ MeV/c^2

\Delta m=1.581\ \cdot\ 10^{25}\ MeV/c^2

Step 3: Determining binding energy of the given nucleus

Binding energy \Delta E_d = \Delta mc^2

\Delta E_d=1.581\ \cdot10^{25}\ MeV/c^2*c^2

\Delta E_d=1.581\cdot\ 10^{25}\ MeV

Therefore, binding energy of the nucleus of the coin \Delta E_d=1.581\cdot\ 10^{25}\ MeV\Delta E_d=1.581\ MeV

Converting MeV to joules

1 MeV = 1.6 * 10^{-13} J

\Delta E_d=1.581\cdot10^{25}\cdot\ 1.6\cdot\ 10^{-13}\ J

\Delta E_d = 2.53 * 10^{12} J

Hence, energy required to separate all the protons and neutrons is \Delta E_d = 2.53 * 10^{12} J